(A) For an adiabatic process,the first law of thermodynamics states that $\Delta U = q + w$. Since $q = 0$,$\Delta U = w$.Given $w = -3\, kJ = -3000\,

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(A) For an adiabatic process,the first law of thermodynamics states that $\Delta U = q + w$. Since $q = 0$,$\Delta U = w$.Given $w = -3\, kJ = -3000\, J$ (work done by the system is negative).We know $\Delta U = n \times C_v \times \Delta T$.Substituting the values: $-3000 = 1 \times 20 \times (T_f - 300)$.$-150 = T_f - 300$.$T_f = 300 - 150 = 150\, K$.

Class 11 Chemistry Thermodynamics First law of thermodynamics

Medium

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27\, ^{\circ}C$. If the work done during the process is $3\, kJ$,then the final temperature of the gas is $(C_v = 20\, J\, K^{-1}\, mol^{-1})$ .....$K$

A
$150$
B
$450$
C
$1750$
D
$225$

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Thermodynamics

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First law of thermodynamics

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EasyMHT CET 2020

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One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27\, ^{\circ}C$. If the work done during the process is $3\, kJ$,then the final temperature of the gas is $(C_v = 20\, J\, K^{-1}\, mol^{-1})$ .....$K$

Similar Questions

$A$ system does $394 \ J$ of work on the surroundings by absorbing $701 \ J$ of heat. What is the change in internal energy of the system (in $J$)?

For $1 \, \text{mol}$ of an ideal gas,isothermal expansion occurs from $1 \, \text{L}$ to $10 \, \text{L}$ in a vacuum at $300 \, \text{K}$. What is the value of $\Delta U$ for this process? $(R = 2 \, \text{cal} \, \text{K}^{-1} \, \text{mol}^{-1})$

“The mass and energy both are conserved in an isolated system”,is the statement of

Calculate the difference between $\Delta H$ and $\Delta U$ for the combustion of $n$-octane at $25^{\circ}C$. (in $kJ$)

For the reaction,$N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$,$\Delta H$ is equal to

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